How do you differentiate $y=a{{x}^{2}}+bx+c$ ?

Hint: Now to differentiate the given function we will use properties of differentiation. Now we know that $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$ , $\dfrac{d\left( c \right)}{dx}=0$ and $\dfrac{d\left( cx \right)}{dx}=c\dfrac{dx}{dx}$ . Hence using this we will first simplify the given differentiation. Now we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ . Hence using this we can easily find the differentiation of the given function.

Complete step-by-step answer:
Now let us first understand the concept of differentiation.
Now the differentiation of a function also known as the derivative of a function is nothing but the rate of change of a function.
Now we know that every function has an independent variable.
So the differentiation of the function gives us rate of change of function per unit change in independent variable. Hence differentiation is defined with respect to a particular variable.
Now differentiation of a function y with respect to x is written as $\dfrac{dy}{dx}$ .
Now consider the given function $a{{x}^{2}}+bx+c$ .
Now let us differentiate the function.
We know that $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$
Hence we have $\dfrac{d\left( a{{x}^{2}}+bx+c \right)}{dx}=\dfrac{d\left( a{{x}^{2}} \right)}{dx}+\dfrac{d\left( bx \right)}{dx}+\dfrac{d\left( c \right)}{dx}$
Now we have that $\dfrac{d\left( c \right)}{dx}=0$ and $\dfrac{d\left( cx \right)}{dx}=c\dfrac{df}{dx}$ where c is a constant
Hence using this we get, $a\dfrac{d\left( {{x}^{2}} \right)}{dx}+b\dfrac{d\left( x \right)}{dx}+0$
Now we have that the differentiation of any function of the form ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ . hence we have,
$\Rightarrow a\left( 2x \right)+b\left( 1 \right)$
Hence the differentiation of the given function is given by $2ax+b$ .

Note: bNow note that the differentiation is rate of change of function. Now here we know that the given function is a quadratic equation in x. Now here we have the differentiation of the function as a function. This is nothing but the equation of slope of the line at point x. Hence we have that differentiation of a curve gives us slope of the curve at point x.